iteration with biscetion method

Hello i need som help with program this code. I have the following parameters in a matrix[1566,1] R, I, Q_min, Q_max, A and d. And i want to solv the eq with the biscetion methode. i want the result to come out in a matriv?
%%%%%%%%%%%%%%%%%%%%%%%%%%Parameter %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
data=xlsread('ledning.xls');
Q_min=data(:,2);
R=data(:,5)/4; % Hydraulisk radius for fuldløbende ledning
I=data(:,4); %I=I_0 dvs. slope
A=(data(:,5)/2).^2*3.14;
M=75;
d=data(:,5);
for i=1:length(data)
Q_max(i,1)=abs(A(i,1)*M*(I(i,1)/100)^(1/2)*R(i,1)^(2/3));
end
for i=1:length(data)
p(i,1)=Q_min(i,1)/Q_max(i,1);
end
%%%%%%%%%%%%%%%%%%%%%%%%Ligning der skal løses %%%%%%%%%%%%%%%%%%%%%%%%%%%
%Ligningen løses via biscetion metoden
f=@(y) (0.46-0.5*cos(3.14*(y/d))+0.004*cos(2*3.14*(y/d)))-(p); %delfyldningsformel
format long
%%%%%%%%%%%%%%%%%%%%%%%%Betingelser til biscetion %%%%%%%%%%%%%%%%%%%%%%%%
eps_abs = 1e-5;
eps_step = 1e-5; %%%%abs og step angiver hvor tæt resultaterne skal være på hinanden
a = 0; % Initial Guess to your function such that f(a)>0.
b = 0.2; % Initial Guess to your function such that f(b)<0.
%%%%%%%%%%%%%%%%%%%%%%%%Påbegynd loop %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
while (b - a >= eps_step || ( abs( f(a) ) >= eps_abs && abs( f(b) ) >= eps_abs ) )
c = (a + b)/2;
if ( f(c) == 0 )
break;
elseif ( f(a)*f(c) < 0 )
b = c;
else
a = c;
end
end

댓글 수: 1

Jan
Jan 2013년 3월 12일
What does the posted code do in opposite to your demands?

댓글을 달려면 로그인하십시오.

답변 (0개)

카테고리

도움말 센터File Exchange에서 Creating and Concatenating Matrices에 대해 자세히 알아보기

질문:

2013년 3월 12일

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by